To solve this problem, we need to determine how many spins are expected before we get a multiple of 3 from a spinner that is divided into 8 sectors labeled 1 through 8.
The multiples of 3 from 1 to 8 are 3 and 6. Therefore, there are 2 favorable outcomes (3 and 6) out of 8 possible outcomes.
The probability \( P \) of spinning a multiple of 3 is given by:
\[ P = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{2}{8} = \frac{1}{4}. \]
Now, we are interested in the number of spins needed for a multiple of 3 to come up 6 times.
Let the number of spins required to get one multiple of 3 be denoted as \( X \). Since the probability of spinning a multiple of 3 is \( \frac{1}{4} \), the expected number of spins \( E(X) \) to get one multiple of 3 can be calculated using the formula for the expectation of a geometric distribution:
\[ E(X) = \frac{1}{P} = \frac{1}{\frac{1}{4}} = 4. \]
Thus, we expect to spin the spinner 4 times to get 1 multiple of 3.
To find the expected number of spins to get 6 multiples of 3, we multiply the expected number of spins to get one multiple by 6:
\[ E(\text{total spins}) = 6 \times E(X) = 6 \times 4 = 24. \]
Therefore, the spinner is expected to have to spin approximately 24 times for a multiple of 3 to be spun 6 times.